MATHS 2 – SCIENCE (Mind Faces)
(Quick Guidance CHEAT SHEET)
Note – Questions given here are indicative and typical. However, in exams, only
these kind of questions will be asked
1. Differentiation
1.1 Definition :
If lim f (x + h) − f (x) exists, then this limits is called is the derivative of f (x) w.r.t.x
h→0 h f (x + h) − f (x)
and is denoted by f ' (x) (x) = lim
h→0 h
Leibnitz’s Notation :
Let y = f (x) stbe a function of x. Let y be a small change in y corresponding to change
x in x, then lim [x ] y
exists,is called the
h→0
y
derivative of y w.r.t.x and is denoted by dx i. e. dx
dy
= lim
dy
[ ]
h→0 x
dy d
Let y = f (x) then di ff.y w.r.t. x is = f (x) = f' (x)
dx dx
Sr. No. Derivatives of Standard Function Derivative of Composite Function
1) d
(k) = 0, k is const.
dx
n n−1
2) d d
xn = nxn − 1 [f (x)] = n [f (x)] f ' (x)
dx dx
3) d 1 −1 d 1 −1
=
dx (x) =
2'
dx x x f ' (x)
f [f (x)]2
4) 1 d 1
d
f (x) = 2 f(x) f ' (x)
dx
=
dx x 2 x
5) d d sin f (x) = cos f (x) f ' (x)
sin x = cos x dx
dx
6) d d cos f (x) = − sin f (x) f ' (x)
cos x = − sin x
dx dx
7) d d
dx tan x = sec x dx tan f (x) = sec f (x) f ' (x)
2 2
8) d d
cot x = − cosec2 x cot f (x) = − cosec2 f (x) f ' (x)
dx dx
Standard - XII : Subject - Mathematics and Statistics (Arts and Science) Part - II : 1
, Sr. No. Derivatives of Standard Function Derivative of Composite Function
9) d d
sec x = sec x tan x sec f (x) = sec (x) tan f (x) f ' (x)
dx dx
10) d d
cosec x = − cosec x cot x cosec f (x)
dx dx
11) d
= ax log a, d
dx a f (x) = a f (x) log a f ' (x), a 1, a
a 1, a 1 dx
12) d d f (x)
e x= e x e = e f (x) f ' (x)
dx dx
13) d 1 d 1
log x, = x log f (x) = f ' (x)
dx dx f (x)
14) d 1 d 1
log f (x) = f ' (x)
dx loga x, = x log a dx a f (x) log a
a 1, a 1 a 1, a 1
Rules of Differentiation : Let u and v are differentiable functions of x.
1) dy dv
If y = u + v, then =
du
+
dx dx dx
dy du dv
2) If y = u − v, then = dx −
dx dx
3) dy
dv du
If y = u . v, then = u dx + v dx
dx
4) dy
If y = k . u, then =k.
du
where k is constant
dx dx
5) dw dydv du
If y = u . v . w, then
= uv dx + uw dx + vw dx
dx
du − dv
6) u dy v u
If y = dx dx
, then =
v dx v2
Standard - XII : Subject - Mathematics and Statistics (Arts and Science) Part - II : 2
,Theorem : If y is differential function of u and u is differential function of x, then y is
dy dy du
differential function of x and = .
dx du dx
Proof : A Let y and u be a small change in y and u corresponding to change x in x
As x → 0, y → 0 and u → 0
dy lim y
y is differential function of u then = u→0 u
du
du lim u
u is differential function of x then dx = x→0
x
y y u
Now, = .
x u x
Taking limit as x→0
lim y lim y u lim y lim u
x→0 = x→0 . = x→0 . x→0
x u x u x
lim y lim u
= u→0 x 0, u→0
. x→0 →
u x
lim y dy du
x→0 = . dx
x du
y is differential function of x and
Here L.H.S. is exists so R.H.S. is exists
dy dy du
= .
dx du dx
To finding the derivative of composite function is called Chain Rule
dy
1) Find If y = etan x
dx
Solution : y = etan x
Differentiate w.r.t.x
dy d
= etan x
dx tan x = e
tan x
sec2 x
dx
dy
dx If y = tan x
2) Find
Solution : y = tan x
, Differentiate w.r.t.x
dy 1 d d
sec2 x dx x
1
= tan x =
dx dx
2 tan x 2 tan x
1 1
sec2 x
2 tan x 2 x
dy
3) Find If y = cot2 (x3)
dx
Solution : y = cot2 (x3)
Differentiate w.r.t.x
dy d d
= 2 cot (x)3
dx [cot (x) ] = 2 cot (x) [− cosec (x) ] dx
3 3 2 3 (x)3
dx
= 2 cot(x)3[−cosec2(x3)](3x2) = −6x2cot(x)3cosec2(x2)
dy
4) Find If y = log (sec ex2)
dx
Solution : y = log (sec ex2) Differentiate w.r.t.x
dy 1 d 1 d
= sec ex2 = sec ex2 tan ex2 ex2
dx sec ex2 dx sec ex2 dx
d 2
= tan ex2 ex2 x = tan ex2 ex2 2x = 2x ex2 tan ex2
dx
5) Find
dy
If y = log 4x [( ) x2 + 2 3
2
dx
x +2 3
] x2 − 1 2 2
[( )
2
x2 + 1 3 = xlog 4 + 2 log x2 + 2
Solution : y = log x −1
2 = log (4 ) + log 2
4x x
x −1 3
( )
x2 − 1
( )
2 2
y = x log 4 + 3 log (x + 1) − 3 log (x2 − 1)
2
Differentiate w.r.t.x
dy d 2 d 2 d
dx = dx x log 4 + 3 dx log (x + 1) − 3 dx log (x − 1)
2 2
2 1 2 1
= log 4 + 3 (x2 + 1) 2x − 3 (x2 − 1) 2x
(Quick Guidance CHEAT SHEET)
Note – Questions given here are indicative and typical. However, in exams, only
these kind of questions will be asked
1. Differentiation
1.1 Definition :
If lim f (x + h) − f (x) exists, then this limits is called is the derivative of f (x) w.r.t.x
h→0 h f (x + h) − f (x)
and is denoted by f ' (x) (x) = lim
h→0 h
Leibnitz’s Notation :
Let y = f (x) stbe a function of x. Let y be a small change in y corresponding to change
x in x, then lim [x ] y
exists,is called the
h→0
y
derivative of y w.r.t.x and is denoted by dx i. e. dx
dy
= lim
dy
[ ]
h→0 x
dy d
Let y = f (x) then di ff.y w.r.t. x is = f (x) = f' (x)
dx dx
Sr. No. Derivatives of Standard Function Derivative of Composite Function
1) d
(k) = 0, k is const.
dx
n n−1
2) d d
xn = nxn − 1 [f (x)] = n [f (x)] f ' (x)
dx dx
3) d 1 −1 d 1 −1
=
dx (x) =
2'
dx x x f ' (x)
f [f (x)]2
4) 1 d 1
d
f (x) = 2 f(x) f ' (x)
dx
=
dx x 2 x
5) d d sin f (x) = cos f (x) f ' (x)
sin x = cos x dx
dx
6) d d cos f (x) = − sin f (x) f ' (x)
cos x = − sin x
dx dx
7) d d
dx tan x = sec x dx tan f (x) = sec f (x) f ' (x)
2 2
8) d d
cot x = − cosec2 x cot f (x) = − cosec2 f (x) f ' (x)
dx dx
Standard - XII : Subject - Mathematics and Statistics (Arts and Science) Part - II : 1
, Sr. No. Derivatives of Standard Function Derivative of Composite Function
9) d d
sec x = sec x tan x sec f (x) = sec (x) tan f (x) f ' (x)
dx dx
10) d d
cosec x = − cosec x cot x cosec f (x)
dx dx
11) d
= ax log a, d
dx a f (x) = a f (x) log a f ' (x), a 1, a
a 1, a 1 dx
12) d d f (x)
e x= e x e = e f (x) f ' (x)
dx dx
13) d 1 d 1
log x, = x log f (x) = f ' (x)
dx dx f (x)
14) d 1 d 1
log f (x) = f ' (x)
dx loga x, = x log a dx a f (x) log a
a 1, a 1 a 1, a 1
Rules of Differentiation : Let u and v are differentiable functions of x.
1) dy dv
If y = u + v, then =
du
+
dx dx dx
dy du dv
2) If y = u − v, then = dx −
dx dx
3) dy
dv du
If y = u . v, then = u dx + v dx
dx
4) dy
If y = k . u, then =k.
du
where k is constant
dx dx
5) dw dydv du
If y = u . v . w, then
= uv dx + uw dx + vw dx
dx
du − dv
6) u dy v u
If y = dx dx
, then =
v dx v2
Standard - XII : Subject - Mathematics and Statistics (Arts and Science) Part - II : 2
,Theorem : If y is differential function of u and u is differential function of x, then y is
dy dy du
differential function of x and = .
dx du dx
Proof : A Let y and u be a small change in y and u corresponding to change x in x
As x → 0, y → 0 and u → 0
dy lim y
y is differential function of u then = u→0 u
du
du lim u
u is differential function of x then dx = x→0
x
y y u
Now, = .
x u x
Taking limit as x→0
lim y lim y u lim y lim u
x→0 = x→0 . = x→0 . x→0
x u x u x
lim y lim u
= u→0 x 0, u→0
. x→0 →
u x
lim y dy du
x→0 = . dx
x du
y is differential function of x and
Here L.H.S. is exists so R.H.S. is exists
dy dy du
= .
dx du dx
To finding the derivative of composite function is called Chain Rule
dy
1) Find If y = etan x
dx
Solution : y = etan x
Differentiate w.r.t.x
dy d
= etan x
dx tan x = e
tan x
sec2 x
dx
dy
dx If y = tan x
2) Find
Solution : y = tan x
, Differentiate w.r.t.x
dy 1 d d
sec2 x dx x
1
= tan x =
dx dx
2 tan x 2 tan x
1 1
sec2 x
2 tan x 2 x
dy
3) Find If y = cot2 (x3)
dx
Solution : y = cot2 (x3)
Differentiate w.r.t.x
dy d d
= 2 cot (x)3
dx [cot (x) ] = 2 cot (x) [− cosec (x) ] dx
3 3 2 3 (x)3
dx
= 2 cot(x)3[−cosec2(x3)](3x2) = −6x2cot(x)3cosec2(x2)
dy
4) Find If y = log (sec ex2)
dx
Solution : y = log (sec ex2) Differentiate w.r.t.x
dy 1 d 1 d
= sec ex2 = sec ex2 tan ex2 ex2
dx sec ex2 dx sec ex2 dx
d 2
= tan ex2 ex2 x = tan ex2 ex2 2x = 2x ex2 tan ex2
dx
5) Find
dy
If y = log 4x [( ) x2 + 2 3
2
dx
x +2 3
] x2 − 1 2 2
[( )
2
x2 + 1 3 = xlog 4 + 2 log x2 + 2
Solution : y = log x −1
2 = log (4 ) + log 2
4x x
x −1 3
( )
x2 − 1
( )
2 2
y = x log 4 + 3 log (x + 1) − 3 log (x2 − 1)
2
Differentiate w.r.t.x
dy d 2 d 2 d
dx = dx x log 4 + 3 dx log (x + 1) − 3 dx log (x − 1)
2 2
2 1 2 1
= log 4 + 3 (x2 + 1) 2x − 3 (x2 − 1) 2x
